Integrand size = 99, antiderivative size = 32 \[ \int \frac {-25+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} \left (-25-50 e^{x^2} x\right )}{1+e^{-4+4 e^{\sqrt [4]{e}}-2 e^{x^2}-2 x}+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} (-2-2 x)+2 x+x^2} \, dx=\frac {25}{1-e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x}+x} \] Output:
25/(x+1-exp(2*exp(exp(1/4))-exp(x^2)-x-2))
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {-25+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} \left (-25-50 e^{x^2} x\right )}{1+e^{-4+4 e^{\sqrt [4]{e}}-2 e^{x^2}-2 x}+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} (-2-2 x)+2 x+x^2} \, dx=\frac {25 e^{2+e^{x^2}+x}}{-e^{2 e^{\sqrt [4]{e}}}+e^{2+e^{x^2}+x} (1+x)} \] Input:
Integrate[(-25 + E^(-2 + 2*E^E^(1/4) - E^x^2 - x)*(-25 - 50*E^x^2*x))/(1 + E^(-4 + 4*E^E^(1/4) - 2*E^x^2 - 2*x) + E^(-2 + 2*E^E^(1/4) - E^x^2 - x)*( -2 - 2*x) + 2*x + x^2),x]
Output:
(25*E^(2 + E^x^2 + x))/(-E^(2*E^E^(1/4)) + E^(2 + E^x^2 + x)*(1 + x))
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{-e^{x^2}-x+2 e^{\sqrt [4]{e}}-2} \left (-50 e^{x^2} x-25\right )-25}{x^2+e^{-2 e^{x^2}-2 x+4 e^{\sqrt [4]{e}}-4}+e^{-e^{x^2}-x+2 e^{\sqrt [4]{e}}-2} (-2 x-2)+2 x+1} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 \left (e^{x^2}+x+2\right )} \left (e^{-e^{x^2}-x+2 e^{\sqrt [4]{e}}-2} \left (-50 e^{x^2} x-25\right )-25\right )}{\left (-e^{e^{x^2}+x+2} x-e^{e^{x^2}+x+2}+e^{2 e^{\sqrt [4]{e}}}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {50 x \exp \left (x^2-e^{x^2}+2 \left (e^{x^2}+x+2\right )-x-2 \left (1-e^{\sqrt [4]{e}}\right )\right )}{\left (-e^{e^{x^2}+x+2} x-e^{e^{x^2}+x+2}+e^{2 e^{\sqrt [4]{e}}}\right )^2}-\frac {25 e^{-e^{x^2}+2 \left (e^{x^2}+x+2\right )-x-2} \left (e^{e^{x^2}+x+2}+e^{2 e^{\sqrt [4]{e}}}\right )}{\left (e^{e^{x^2}+x+2} x+e^{e^{x^2}+x+2}-e^{2 e^{\sqrt [4]{e}}}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -50 \int \frac {\exp \left (x^2-x-e^{x^2}+2 \left (x+e^{x^2}+2\right )-2 \left (1-e^{\sqrt [4]{e}}\right )\right ) x}{\left (-e^{x+e^{x^2}+2} x-e^{x+e^{x^2}+2}+e^{2 e^{\sqrt [4]{e}}}\right )^2}dx-25 \int \frac {e^{-x-e^{x^2}+2 \left (x+e^{x^2}+2\right )-2 \left (1-e^{\sqrt [4]{e}}\right )}}{\left (-e^{x+e^{x^2}+2} x-e^{x+e^{x^2}+2}+e^{2 e^{\sqrt [4]{e}}}\right )^2}dx-25 \int \frac {e^{-x-e^{x^2}+2 \left (x+e^{x^2}+2\right )-2 \left (1-e^{\sqrt [4]{e}}\right )}}{(x+1) \left (e^{x+e^{x^2}+2} x+e^{x+e^{x^2}+2}-e^{2 e^{\sqrt [4]{e}}}\right )^2}dx-25 \int \frac {e^{-x-e^{x^2}+2 \left (x+e^{x^2}+2\right )-2}}{(x+1) \left (e^{x+e^{x^2}+2} x+e^{x+e^{x^2}+2}-e^{2 e^{\sqrt [4]{e}}}\right )}dx\) |
Input:
Int[(-25 + E^(-2 + 2*E^E^(1/4) - E^x^2 - x)*(-25 - 50*E^x^2*x))/(1 + E^(-4 + 4*E^E^(1/4) - 2*E^x^2 - 2*x) + E^(-2 + 2*E^E^(1/4) - E^x^2 - x)*(-2 - 2 *x) + 2*x + x^2),x]
Output:
$Aborted
Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84
method | result | size |
norman | \(\frac {25}{x +1-{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{\frac {1}{4}}}-{\mathrm e}^{x^{2}}-x -2}}\) | \(27\) |
risch | \(\frac {25}{x +1-{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{\frac {1}{4}}}-{\mathrm e}^{x^{2}}-x -2}}\) | \(27\) |
parallelrisch | \(\frac {25}{x +1-{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{\frac {1}{4}}}-{\mathrm e}^{x^{2}}-x -2}}\) | \(27\) |
Input:
int(((-50*exp(x^2)*x-25)*exp(2*exp(exp(1/4))-exp(x^2)-x-2)-25)/(exp(2*exp( exp(1/4))-exp(x^2)-x-2)^2+(-2-2*x)*exp(2*exp(exp(1/4))-exp(x^2)-x-2)+x^2+2 *x+1),x,method=_RETURNVERBOSE)
Output:
25/(x+1-exp(2*exp(exp(1/4))-exp(x^2)-x-2))
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-25+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} \left (-25-50 e^{x^2} x\right )}{1+e^{-4+4 e^{\sqrt [4]{e}}-2 e^{x^2}-2 x}+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} (-2-2 x)+2 x+x^2} \, dx=\frac {25}{x - e^{\left (-x - e^{\left (x^{2}\right )} + 2 \, e^{\left (e^{\frac {1}{4}}\right )} - 2\right )} + 1} \] Input:
integrate(((-50*exp(x^2)*x-25)*exp(2*exp(exp(1/4))-exp(x^2)-x-2)-25)/(exp( 2*exp(exp(1/4))-exp(x^2)-x-2)^2+(-2-2*x)*exp(2*exp(exp(1/4))-exp(x^2)-x-2) +x^2+2*x+1),x, algorithm="fricas")
Output:
25/(x - e^(-x - e^(x^2) + 2*e^(e^(1/4)) - 2) + 1)
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {-25+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} \left (-25-50 e^{x^2} x\right )}{1+e^{-4+4 e^{\sqrt [4]{e}}-2 e^{x^2}-2 x}+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} (-2-2 x)+2 x+x^2} \, dx=- \frac {25}{- x + e^{- x - e^{x^{2}} - 2 + 2 e^{e^{\frac {1}{4}}}} - 1} \] Input:
integrate(((-50*exp(x**2)*x-25)*exp(2*exp(exp(1/4))-exp(x**2)-x-2)-25)/(ex p(2*exp(exp(1/4))-exp(x**2)-x-2)**2+(-2-2*x)*exp(2*exp(exp(1/4))-exp(x**2) -x-2)+x**2+2*x+1),x)
Output:
-25/(-x + exp(-x - exp(x**2) - 2 + 2*exp(exp(1/4))) - 1)
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {-25+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} \left (-25-50 e^{x^2} x\right )}{1+e^{-4+4 e^{\sqrt [4]{e}}-2 e^{x^2}-2 x}+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} (-2-2 x)+2 x+x^2} \, dx=\frac {25 \, e^{\left (x + e^{\left (x^{2}\right )} + 2\right )}}{{\left (x e^{2} + e^{2}\right )} e^{\left (x + e^{\left (x^{2}\right )}\right )} - e^{\left (2 \, e^{\left (e^{\frac {1}{4}}\right )}\right )}} \] Input:
integrate(((-50*exp(x^2)*x-25)*exp(2*exp(exp(1/4))-exp(x^2)-x-2)-25)/(exp( 2*exp(exp(1/4))-exp(x^2)-x-2)^2+(-2-2*x)*exp(2*exp(exp(1/4))-exp(x^2)-x-2) +x^2+2*x+1),x, algorithm="maxima")
Output:
25*e^(x + e^(x^2) + 2)/((x*e^2 + e^2)*e^(x + e^(x^2)) - e^(2*e^(e^(1/4))))
Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-25+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} \left (-25-50 e^{x^2} x\right )}{1+e^{-4+4 e^{\sqrt [4]{e}}-2 e^{x^2}-2 x}+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} (-2-2 x)+2 x+x^2} \, dx=\frac {25}{x - e^{\left (-x - e^{\left (x^{2}\right )} + 2 \, e^{\left (e^{\frac {1}{4}}\right )} - 2\right )} + 1} \] Input:
integrate(((-50*exp(x^2)*x-25)*exp(2*exp(exp(1/4))-exp(x^2)-x-2)-25)/(exp( 2*exp(exp(1/4))-exp(x^2)-x-2)^2+(-2-2*x)*exp(2*exp(exp(1/4))-exp(x^2)-x-2) +x^2+2*x+1),x, algorithm="giac")
Output:
25/(x - e^(-x - e^(x^2) + 2*e^(e^(1/4)) - 2) + 1)
Time = 3.98 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-25+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} \left (-25-50 e^{x^2} x\right )}{1+e^{-4+4 e^{\sqrt [4]{e}}-2 e^{x^2}-2 x}+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} (-2-2 x)+2 x+x^2} \, dx=\frac {25}{x-{\mathrm {e}}^{-{\mathrm {e}}^{x^2}}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{{\mathrm {e}}^{1/4}}}+1} \] Input:
int(-(exp(2*exp(exp(1/4)) - exp(x^2) - x - 2)*(50*x*exp(x^2) + 25) + 25)/( 2*x + exp(4*exp(exp(1/4)) - 2*exp(x^2) - 2*x - 4) + x^2 - exp(2*exp(exp(1/ 4)) - exp(x^2) - x - 2)*(2*x + 2) + 1),x)
Output:
25/(x - exp(-exp(x^2))*exp(-x)*exp(-2)*exp(2*exp(exp(1/4))) + 1)
Time = 124.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.72 \[ \int \frac {-25+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} \left (-25-50 e^{x^2} x\right )}{1+e^{-4+4 e^{\sqrt [4]{e}}-2 e^{x^2}-2 x}+e^{-2+2 e^{\sqrt [4]{e}}-e^{x^2}-x} (-2-2 x)+2 x+x^2} \, dx=-\frac {25 e^{e^{x^{2}}+x} e^{2}}{e^{2 e^{e^{\frac {1}{4}}}}-e^{e^{x^{2}}+x} e^{2} x -e^{e^{x^{2}}+x} e^{2}} \] Input:
int(((-50*exp(x^2)*x-25)*exp(2*exp(exp(1/4))-exp(x^2)-x-2)-25)/(exp(2*exp( exp(1/4))-exp(x^2)-x-2)^2+(-2-2*x)*exp(2*exp(exp(1/4))-exp(x^2)-x-2)+x^2+2 *x+1),x)
Output:
( - 25*e**(e**(x**2) + x)*e**2)/(e**(2*e**(e**(1/4))) - e**(e**(x**2) + x) *e**2*x - e**(e**(x**2) + x)*e**2)